By J.-L. Colliot-Thelene, M. Coste, L. Mahe, M.-F. Roy

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**Extra resources for Geometrie Algebrique Reelle et Formes Quadratiques**

**Example text**

Then and 1, I const jln(1 - T)], {(I - 2 ) 3 - n , n = 2, n = 3, n > 3. 50 11. The Basic Examples On the other hand, for z > $, we have 2 const(1 - T ) ' - ~ . Thus, by Rayleigh's theorem, we have 1 - tanh6/2, - tanh 6/2)2Iln(1 - tanh 6/2)1, n > 2, n = 3, n > 3, for large 6, and the theorem follows. THEOREM6. Let K > 0, that is, MIK = S n ( l / f i ) Then . (51) and p(6) 2 (52) for all 6 E (0,n / J i ) . (n - 1 ) ~ PROOF:Here sK(t) = (sinfit)/&, and, for our convenience, we shall set K in our coordinate system is given by ds2 = cK(t) = cosfit, = 1.

Represent F in the geodesic spherical coordinates by (28). Then for each fixed t E [O, S] we have the decomposition in L2(S"-'), m fk 5 ) = c 4t)G,(5)7 l=O where GI is an eigenfunction of v 1 on 9-',with L2(S"-')-norm equal to 1. Thus m P (39) ', where we are letting d A denote Riemannian measure on S"- and a,@)= Jsn- * OGl(5)dA(5). f ( t 9 Therefore we have -V&) = J sn-1 - S,-. 1 f ( 4 5 ) oc1(5)dA(o (U

The Laplacian Let G,, . , Gk, Gk+ . . be nodal domains of define h =I : { 4 . For each j = 1, . ,k on Gj, on @ - G j . Gj One then obtains, as above, the existence of a nontrivial function k satisfying 0 = = ... = (f; A- 1). One verifies that h E &(M) for each j = 1, . , k. ,, But then the maximum principle (cf. 11) implies that f vanishes identically on M-a contradiction. Before proceeding to case (ii) we first note an immediate consequence of the nodal domain theorem. COROLLARY 2. $1 always has constant sign; 1, has multiplicity equal to 1; and 42 has precisely 2 nodal domains.